(w+9)^3+56=0 write were w i a real number write answer in simplified radical form
write b^2/3 as a radical expression
for the second one you have b^2 under a radical sign with an index of 3 (or root of 3)
You have some typos in the first one.. but if I am a reading it correctly.
(w+9)^3 = -56
take the cube root of both sides
w + 9 = cube root of (-56)
subtract 9.. the 9 will not be under the root.
To solve the equation (w+9)^3 + 56 = 0, we can follow these steps:
Step 1: Start by subtracting 56 from both sides of the equation:
(w+9)^3 = -56
Step 2: Take the cube root of both sides to eliminate the exponent:
∛((w+9)^3) = ∛(-56)
Step 3: Simplify the cube root on the left side:
w + 9 = -∛56
Step 4: Subtract 9 from both sides to isolate w:
w = -9 - ∛56
So, the solution to the equation (w+9)^3 + 56 = 0, where w is a real number, is w = -9 - ∛56.
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To express b^(2/3) as a radical expression, follow these steps:
Step 1: Recall that when a number is raised to the power of a fraction, it is equivalent to taking the nth root of that number raised to the power of the numerator.
Step 2: The exponent 2/3 can be written as the power of a radical expression:
b^(2/3) = (√b)^2
So, b^(2/3) can be expressed as (√b)^2, where b is a real number.